RoseHulman
Undergraduate
Mathematics
Journal
Points of Ninth Order on Cubic
Curves
Leah Balay-Wilson
a
Taylor Brysiewicz
Volume 15, No. 1, Spring 2014
Sponsored by
Rose-Hulman Institute of Technology
Department of Mathematics
Terre Haute, IN 47803
Email: mathjournal@rose-hulman.edu
a Smith
http://www.rose-hulman.edu/mathjournal
b Northern
College
Illinois University
b
Rose-Hulman Undergraduate Mathematics Journal
Volume 15, No. 1, Spring 2014
Points of Ninth Order on Cubic Curves
Leah Balay-Wilson
Taylor Brysiewicz
Abstract. In this paper we geometrically provide a necessary and sufficient condition for points on a cubic to be associated with an infinite family of other cubics
who have nine-pointic contact at that point. We then provide a parameterization of
the family of cubics with nine-pointic contact at that point, based on the osculating
quadratic.
Acknowledgements: This research took place at the University of Wisconsin – Stout
REU 2013, which was sponsored by NSF grant DMS 1062403. The authors wish to acknowledge the support of the University of Wisconsin – Stout. In particular, Dr. Seth
Dutter, for all of his guidance and support throughout this project.
Page 2
1
RHIT Undergrad. Math. J., Vol. 15, No. 1
Introduction
Algebraic geometry studies geometry of sets defined by the common solutions to polynomial
equations. One of the classic topics of study is the intersections between algebraic curves
in the plane. Algebraic geometers quantify intersection between two curves by computing
the intersection multiplicity of the curves at that point. Curves that intersect each other
to a high multiplicity at a point will be good local approximations for each other. In the
simplest sense, osculating curves are the best local approximations to a curve at a point and
correspondingly, extactic points are the points on a curve where even an osculating curve
intersects to a greater multiplicity than expected. The product of the degree of two curves
provides an upper bound for the intersection multiplicity between them at any given point
so investigating points on a curve at which that bound is able to be acheived is of particular
interest.
The study of osculating curves was of interest to Cayley and Salmon in the 1800’s.
Cayley was interested in finding the sextactic points of a curve, the points on a curve such
that the curve intersects a conic with multiplicity six at that point. Salmon was interested
in something slightly different: cubic curves that intersect other cubic curves to high degree.
This is the phenomena we are focused on in this paper. What Salmon did was count the
points on a cubic at which it is possible to have another cubic intersect it with multiplicity
9, or equivalently, intersect at exactly one point. He called these the points of nine-pointic
contact. He determined that there are 81 of these points on any cubic, including the 9
inflection points. He also supplied a simple equation for finding these points. These points
and their defining equation were further examined by A.S. Hart in 1875.
Other research, including that of Halphen in his 1876 thesis, focuses on coincidence points.
The concept of these points is not quite the same as that of the points of nine-pointic contact.
However, both of these sets of points are related to the construction of infinite families of
cubic curves that intersect to degree nine at a point.
Though this topic possesses a rich history, research on it continues. While Cayley focused
on finding the sextactic points of a plane curve of arbitrary degree, Kamel and Farahat,
in 2012, looked closely at the total sextactic points of quartic plane curves in [4]. Their
investigation centered around investigating the points on a quartic curve at which there
exists a quadratic curve that intersects it completely at that point, which is to say that they
intersect with multiplicity 8 at that point and nowhere else.
The points of nine-pointic contact investigated by Salmon give rise to infinite families
of cubic curves that intersect to degree nine at that point. Examining which cubic curves
fit into which of these families is an interesting and complicated question. We provide
insight into the answer by demonstrating the geometric conditions required for a point to
be a point of nine-pointic contact and describing how these conditions relate to osculating
curves. The purpose of this paper is to build the infinite family of cubics that intersect a
smooth cubic curve to degree nine at a non-flex point directly from the theory of osculating
curves. Furthermore, we show that these cubic curves are in bijection with the points on the
osculating conic, thus showing that this family is parametrizable.
RHIT Undergrad. Math. J., Vol. 15, No. 1
2
Page 3
Preliminaries
We will assume a basic familiarity of Algebraic Geometry. For the reader who is new to the
subject the authors recommend “Algebraic Curves” by Fulton [3]. However, as our proofs
depend upon them, we will give a brief overview of the necessary definitions and theorems.
By convention, for a polynomial f , the notation V(f ) will denote the vanishing of f , or
equivalently the variety defined by f .
2.1
Intersection Multiplicity
Definition 2.1. Let C, D ∈ C[x, y] and p be a maximal ideal. We define the intersection
multiplicity of C and D at p to be
Ip (C, D) = dimC C[x, y]p /hC, Di,
where the dimension is taken to be the dimension as a C vector space.
We can extend this definition to the intersection multiplicity of two projective curves at
a point p by selecting an affine chart containing p and then applying the above definition
to the defining equations of the given curves at the maximal ideal associated to p. This is
independent of the choice of affine chart and defining equations. By abuse of language we
will speak of the intersection multiplicity of two curves. Additionally we recall the following
properties of intersection multiplicity.
Corollary 2.2. Let C, D, E ∈ C[x, y], and α ∈ C∗ , then intersection multiplicity satisfies
the following properties:
i) Ip (C, D) = Ip (D, C)
ii) Ip (C, C) = ∞
iii) Ip (CD, E) = Ip (C, E) + Ip (D, E)
iv) Ip (αC, D) = Ip (C, D)
v) Ip (C, D) = 0 ⇐⇒ p 6= V(C) ∩ V(D)
vi) Ip (C, D) = Ip (C, D + EC)
vii) I(0,0) (x, y) = 1
Proof. See Fulton [3], Chapter 3, Section 3, Theorem 3.
RHIT Undergrad. Math. J., Vol. 15, No. 1
Page 4
Lemma 2.3. Let C, D, E ∈ C[x, y] and Ip (C, D) = n and Ip (D, E) = m. If V(D) is smooth
at p, then Ip (C, E) ≥ min{n, m}.
Proof. We may assume that p ∈ V(D) because if p is not on the curve V(D) then Ip (C, D) = 0
and Ip (D, E) = 0 so the lemma follows trivially.
Since V(D) is smooth, by [3], Chapter 3, Section 2, Thorem 1, the local ring C[x, y]p /hDi
is a discrete valuation ring. Let t be a local parameter. Next, we consider C and E as
elements of this local ring. Since Ip (C, D) = n and Ip (D, E) = m, we can write C = tn α
and E = tm β for some α and β which are rational functions that are not contained in the
maximal ideal p. Without a loss of generality, we assume n ≤ m, so
C[x, y]p /(C, D, E) = C[x, y]p /hC, Di.
We also have a surjective morphism C[x, y]p /hC, Ei → C[x, y]p /hC, D, Ei, so transitively
there is a surjective morphism C[x, y]p /hC, Ei → C[x, y]p /hC, Di. Thus,
dimC (C[x, y]p /hC, Ei) ≥ dimC (C[x, y]p /hC, Di)
which is the minimum of n and m by assumption.
Lemma 2.4. Intersection multiplicity is invariant under projective linear transformations.
Proof. For details, see [3] page 37.
In affine space, it can be difficult to determine the number of times two curves intersect.
For example, a parabola may intersect a line at two distinct points or only at one distinct
point. However, in projective space we have a definitive answer as to how many times two
curves intersect, given by the following theorem.
Theorem 2.5 (Bézout’s Theorem). Let C, D ⊂ P2C be projective curves of degree m and n.
If C and D share no common components, then the number of points of intersection of C
and D, counting multiplicity, is mn.
Proof. For details concerning the proof of Bézout’s Theorem, see [2] Chapter 8, Section 7,
Theorem 10.
RHIT Undergrad. Math. J., Vol. 15, No. 1
Page 5
Figure 1
3
3
2
2
1
1
0
-1
0
-2
-3
-3
-2
2
-1
0
1
2
2
3
2
(a) V(x − y − 2),V(x + 5y − 5)
-1
-3
-2
-1
1
0
2
3
2
(b) V(y − x ),V(y)
Example 2.6. Consider the affine curves V(x2 − y − 2) and V(x2 + 5y 2 − 5) as depicted
in Figure 1 (a). Since the polynomials x2 − y − 2 and x2 + 5y 2 − 5 are both of second
degree, Bézout’s Theorem implies that there are a total of 4 intersection points including
multiplicity. As shown in the plots of these curves, one can see that there are indeed four
intersection points.
Example 2.7. Recall that Bézout’s Theorem counts multiplicity. In Figure 1 (b). The line
V(y) lies tangent to the parabola V(y − x2 ). We will see in section 2.2 that this tangency
implies an intersection multiplicity of 2 at the point, which must account for all of the points
of intersection since y − x2 is of second degree and y is of first degree.
Bézout’s Theorem is essential to understanding intersection theory and it will play a key
role in many of our proofs. For example, the following lemma follows nicely from Bézout’s
Theorem.
Lemma 2.8. Any two pairs of distinct lines in P2C are projectively equivalent.
Proof. By Bézout’s Theorem, we know that any two lines will intersect precisely once. Let
p0 be that point of intersection and let p1 and p2 be points distinct from p0 on each line
respectively. By Theorem 3.4 of [1], there is a projective change of coordinates that takes
p0 , p1 , p2 to [0 : 0 : 1], [1 : 0 : 1], and [0 : 1 : 1] respectively. Thus, since every pair of distinct
lines is projectively equivalent to the union of the lines x = 0 and y = 0, and because a
projective change of coordinates is invertible by definition, we conclude that any two pairs
of distinct lines are projectively equivalent.
RHIT Undergrad. Math. J., Vol. 15, No. 1
Page 6
2.2
Osculating Curves
Let C ∈ C[x, y]. When considering a point p on V(C), one can construct the tangent line
of V(C) at p. When p is a smooth point the tangent line has the same slope as V(C) at p
and the intersection multiplicity of V(C) and the tangent line is at least 2 at the point p. In
order to give a definition of tangent lines that extends to singular points we will first need a
lemma on homogeneous polynomials in 2 variables.
Lemma 2.9. If C ∈ C[x, y] is a homogeneous polynomial of degree n, then C factors into n
linear homogeneous terms.
Proof. Since C is homogeneous in x and y of degree n, we may write
C=
n
X
ai xi y n−i .
i=0
Suppose that y - C, then an 6= 0. We divide through by y n , so that
i
n
X
x
C
=
a
i
yn
y
i=1
is degree n in x/y. By the Fundamental Theorem of Algebra there exists an a ∈ C and some
αi ∈ C for i = 1, . . . , n such that
n Y
x
C
=a
− αi .
yn
y
i=1
Multiplying through by y n yields
n
Y
C = a (x − αi y).
i=1
In the case that y | C, we may write C = y · C 0 where k is the greatest positive integer such
that y k | C. Then by the previous reasoning, C 0 is a product of n − k linear factors so C is
a product of n linear factors where k of them are the polynomial y.
k
While tangent lines are commonly understood analytically, it will be useful for us to
define a tangent line of a curve at a point formally instead.
Definition 2.10. Let C be a homogeneous polynomial defining a projective curve of degree
n passing through the point [0 : 0 : 1]. We can write C as
C = Lm Z n−m + Lm+1 Z n−m−1 + · · · + Ln
where each Li ∈ C[X, Y ] is a homogeneous polynomial of degree i and m ≥ 1. By Lemma
2.9 we can factor Lm to get
m
Y
Lm =
Ti .
i=1
The line defined by each Ti is said to be tangent to the curve defined by C at [0 : 0 : 1].
RHIT Undergrad. Math. J., Vol. 15, No. 1
Page 7
Remark 2.11. When working in affine coordinates we would have a similar decomposition
C = Lm + Lm−1 + · · · + Ln ,
where Li ∈ C[x, y] is a homogeneous polynomial of degree i. The tangent lines at [0 : 0 : 1]
would be given by the factors of Lm .
We can extend our definition of tangent lines to any appoint p by applying a projective
linear transformation that takes p to [0 : 0 : 1], finding the tangent lines, and then applying
the inverse linear transformation. The resulting lines will be independent of the choice of
projective linear transformation.
Example 2.12. Consider the curve defined by the polynomial C = x − 3xy − y + x2 . We
can group the terms by degree and see that C = (x − y) + (x2 − 3xy). Then by the definition
of tangent line, the line defined by V(x − y) is a tangent line of V(C) at (0, 0) in affine
coordinates. This is the same result one would obtain using implicit differentiation.
Definition 2.13. Let C = L2 + L3 ∈ C[x, y] define an irreducible cubic curve where each
Li is a homogeneous polynomial of degree i and V(C) singular at (0, 0). If L2 = T1 T2 where
V(T1 ) 6= V(T2 ) then we say that (0, 0) is a node of V(C). If V(T1 ) = V(T2 ) then we say that
(0, 0) is a cusp of V(C).
Example 2.14. Let D = x3 − 2xy 2 + x2 − y 2 define a curve. Again, we may group the terms
by degree to see that D = (x2 − y 2 ) + (x3 − 2xy 2 ). Then the grouping of lowest degree is
(x2 − y 2 ). Following the definition of a tangent line, since x2 − y 2 = (x + y)(x − y), we see
that the lines V(x + y) and V(x − y) are the tangent lines of V(D) at (0, 0). Furthermore,
since there were no first degree terms of g(x, y) and the second degree terms decomposed
into distinct linear factors, we see that V(g) has a node at (0, 0).
Remark 2.15. There are precisely two distinct tangent lines of a cubic at a node and only
one tangent line of a cubic at a cusp.
Figure 2 shows the tangent lines on cubic curves at a smooth point, node, and a cusp.
Figure 2: Tangent Lines
3
1
1
0.5
0.5
0
0
-0.5
-0.5
2
1
0
-1
-2
-3
-3
-2
-1
0
2
1
2
3
3
(a) V(x − y + x + xy − y ),
V(x − y)
-1
-1
-0.5
0
1
0.5
3
3
(b) V(xy + x + y ),
V(x), V(y)
-1
-1
-0.5
0
0.5
(c) V(y 2 − x3 ),
V(y)
1
RHIT Undergrad. Math. J., Vol. 15, No. 1
Page 8
At smooth points we can consider tangent lines to be the best linear approximation to
the curve. We can similarly consider the best approximation by a quadratic curve at a given
point. Such a quadratic curve is said to be osculating at the given point. Formally, we define
the osculating quadratic as follows.
Definition 2.16. Let C, Q ∈ C[x, y] where deg(C) ≥ 3, deg(Q) = 2, and p ∈ V(C). We say
that V(Q) is an osculating quadratic of V(C) at p if Ip (Q, C) ≥ 5.
Example 2.17. Figure 3 depicts the osculating quadratic of a particular smooth curve at
the point (0, 0).
1
0.5
0
-0.5
-0.5
0
0.5
1
Figure 3: V(−y 2 + x2 y + x3 − 2x2 + x), V(−y 2 − 2x2 + x)
Definition 2.18. Let C ∈ C[x, y] define an irreducible affine quadratic curve. We say that
V(C) is a parabola if V(C) intersects the line at infinity at one distinct point with intersection
multiplicity 2.
Definition 2.19. Let C ∈ C[x, y] define an irreducible curve that is smooth at p. We say p
is a flex, or a flex point, of V(C) if there exists a line that intersects V(C) at p with degree
3 or more.
Example 2.20. Consider C = y − x3 ∈ C[x, y]. To see that p = (0, 0) is a flex point, observe
that Ip (y, y − x3 ) = Ip (y, x3 ) = 3 so the line V(y) intersects V(C) at p with degree 3. This
example is depicted in Figure 4 with V(y − x3 ) shown in blue and the V(y) shown in red.
RHIT Undergrad. Math. J., Vol. 15, No. 1
Page 9
1
0.5
0
-0.5
-1
-1
-0.5
0
0.5
1
Figure 4: V(y − x3 ), V(y)
Definition 2.21. Let C ∈ C[x, y] define an irreducible curve that is smooth at p. We say p
is a sextactic point of V(C) if there exists a quadratic curve that intersects V(C) at p with
degree 6 or more.
Lemma 2.22. Let C ∈ C[X, Y, Z] be a homogeneous polynomial defining a cubic curve that
is smooth at a point p ∈ V(C) ⊂ P2C . If p is not a flex of V(C), then there exists a projective
linear transformation taking p to [0 : 0 : 1] and V(C) to a curve defined by
C(X, Y, Z) = Z 2 X + ZY 2 + G3 (X, Y )
where G3 (X, Y ) is a homogeneous polynomial of degree 3 depending only upon X and Y .
Proof. By Lemma 2.8, after a linear change of coordinates we may assume that p = [0 : 0 : 1]
and that the tangent line to the projective curve associated to C is given by X = 0. After
which we can write C as
C = Z 2 L1 (X, Y ) + ZL2 (X, Y ) + L3 (X, Y )
where each Li (X, Y ) is homogeneous of degree i. Since our curve is smooth at [0 : 0 : 1] with
tangent line defined by X = 0 we know that L1 (X, Y ) is a non-zero multiple of X. After
scaling by a nonzero constant we get a new equation
C 0 = Z 2 X + Z(aX 2 + bXY + cY 2 ) + L03 (X, Y ).
If c = 0 then the intersection multiplicity with the line X = 0 would be 3 which would
contradict our assumption that p is not a flex. Therefore we may assume that c 6= 0. Our
next step is to perform the linear change of variables sending Z to Z − a2 X − 2b Y and fixing
X and Y . After making this substitution we are left with
(aX + bY )(aX 2 + bXY + 2cY 2 )
+ L03 (X, Y ).
4
√
√
Finally we scale the variable Y by 1/ c for some choice of c to get the desired form.
Each step of this construction is an invertible linear change of coordinates, therefore their
composition is a projective linear transformation.
Z 2 X + Z(cY 2 ) −
RHIT Undergrad. Math. J., Vol. 15, No. 1
Page 10
Remark 2.23. Note that restricting curves to affine coordinates after projectively transforming them will not change the intersection multiplicity of the two curves at that point.
Thus in the following lemma, we will homogenize the cubic polynomial given, projectively
change the curve into the simpler form described in Lemma 2.22 and then restrict this curve
to the affine plane.
Lemma 2.24. Let C ∈ C[x, y] be the defining polynomial for the cubic curve V(C) and p a
smooth non-flex point of V(C). An osculating conic of V(C) at p exists.
Proof. Suppose that we have homogenized C, applied the projective linear transformations
described in Lemma 2.22 and then restricted the resulting curve, V(C), to affine coordinates.
Our defining equation can then be written as
C = x + y 2 + f x3 + gx2 y + hxy 2 + iy 3
for some f, g, h, i ∈ C. To prove our claim, we will explicitly show that the polynomial
Q = (i2 − h)x2 − ixy + y 2 + x defines an osculating conic of V(C) at (0, 0).
Note that
C − (iy + (i2 + h)x + 1)Q = x2 [(2i3 + 2hi + g)y + (i4 + 2hi2 + h2 + f )x].
Therefore we have
Ip (C, Q) = Ip (x2 , Q) + Ip ((2i3 + 2hi + g)y + (i4 + 2hi2 + h2 + f )x, Q).
Since V(x) is tangent to V(Q) at p, Ip (x, Q) = 2 and Ip (x2 , Q) = 4. On the other hand
(2i3 + 2hi + g)y + (i4 + 2hi2 + h2 + f )x and Q both vanish at p, so
Ip ((2i3 + 2hi + g)y + (i4 + 2hi2 + h2 + f )x, Q) ≥ 1.
In particular Ip (C, Q) ≥ 5. Note that Q must be irreducible, since otherwise it would
decompose into two lines, at least of one which would have to intersect C to degree 3 at p
which would contradict our supposition that p is not a flex point.
Lemma 2.25. Every irreducible nodal cubic has at least two irreducible osculating quadratics
at the node.
Proof. Let N ∈ C[x, y] define a nodal cubic at the point p = (0, 0). By Lemma 2.8, we may
projectively change the two tangent lines of V(N ) to be the lines V(x) and V(y) so after
scaling by a factor, we may suppose that
N = xy + f x3 + gx2 y + hxy 2 + iy 3
for some f, g, h, i ∈ C. Note that the constants f and i cannot be zero since otherwise N
would be reducible. A brief calculation will show that Q1 = x + gx2 + hxy + iy 2 intersects
RHIT Undergrad. Math. J., Vol. 15, No. 1
Page 11
N with multiplicity 6 at p.
Ip (N, Q1 ) = Ip (xy + f x3 + gx2 y + hxy 2 + iy 3 , x + gx2 + hxy + iy 2 )
= Ip (f x3 , x + gx2 + hxy + iy 2 )
= 3Ip (x, x + gx2 + hxy + iy 2 )
= 3Ip (x, iy 2 )
= 6.
By symmetry, the quadratic Q2 = y + f x2 + gxy + hy 2 also intersects N with multiplicity
6 at p. To see that Q1 and Q2 must be irreducible, observe that for a reducible quadratic
to intersect a cubic curve at a point with multiplicity 6, it must be the product of two lines
that intersect the cubic curve at that point with multiplicity 3 each. The only lines that
have that property are V(x) and V(y). Since f and i are not zero, we notice that neither Q1
nor Q2 have a factor of x or y. Hence, Q1 and Q2 cannot be the product of two linear terms
and thus are irreducible.
Lemma 2.26. Let C ∈ C[x, y] be an irreducible polynomial. If p ∈ V(C) is smooth, then
the tangent line of V(C) at p is unique and the osculating quadratic of V(C) at p is unique.
Proof. By the definition of smooth, C can be written as C = L1 (x, y)+L2 (x, y)+· · ·+Ln (x, y)
where Li ∈ C[x, y] is homogeneous of degree i and L1 is not identically zero. By definition
of tangent line, V(L1 (x, y)) is the tangent line of V(C).
Similarly, suppose that the polynomials Q, Q0 ∈ C[x, y] both define osculating quadratics
of V(C) at p. It follows that Ip (C, Q) ≥ 5 and Ip (C, Q0 ) ≥ 5 so by Lemma 2.3 we have that
Ip (Q, Q0 ) ≥ 5. However, by Bézout’s Theorem, we know that V(Q) and V(Q0 ) intersect at
precisely 4 points including multiplicity, so they cannot intersect each other with degree 5
at p.
Remark 2.27. Recall that a flex point of a curve is a smooth point such that there exists a
line that intersects the curve at that point with multiplicity 3. Since osculating quadratics
are unique at a smooth point, the only osculating quadratic at a flex point is given by
an equation for the tangent line squared. This double line will in fact intersect the curve
with multiplicity 6 at the flex point. In particular, this means that there is no irreducible
osculating quadratic of a curve at a flex point.
Page 12
3
RHIT Undergrad. Math. J., Vol. 15, No. 1
Families of Cubics with Ninth Order Intersections at
A Point
Theorem 3.1. Let D ∈ C[x, y] define an irreducible cubic curve through the point p. If there
exists an irreducible quadratic polynomial Q, such that Ip (D, Q) ≥ 5, then we can write
D = T 2 l1 + Ql2
where T defines the tangent line to V(Q) at p and l1 and l2 are linear polynomials.
Proof. Since intersection multiplicity is invariant under a projective change of coordinates, it
will suffice for our purposes to consider only the case in which p = (0, 0). After a projective
change of coordinates followed by localization to affine coordinates, we may suppose that
Q(x, y) = y 2 − x, from which it follows that T (x, y) = x defines a tangent line to Q(x, y) at
p.
Since Ip (D, Q) ≥ 5, it follows that
D(y 2 , y) = ay 6 + by 5 .
Constructing the polynomial g(x, y) = D(x, y) − (ax3 + bx2 y), we observe that
g(y 2 , y) = D(y 2 , y) − ay 6 + by 5 = 0.
Since g(x, y) is identically zero in the polynomial ring C[x, y]/hy 2 − xi, we conclude that
g(x, y) ∈ hy 2 − xi and thus y 2 − x divides the polynomial g(x, y). Using this divisibility
relation, we can write
D(x, y) = ax3 + bx2 y + g(x, y)
= x2 (ax + by) + (y 2 − x)l2
= T 2 l1 + Ql2 ,
where l1 and l2 are linear polynomials in C[x, y]. Note that l1 = ax+by vanishes at p = (0, 0).
This will be used in the next theorem.
We have geometric interpretations for V(T ) and V(Q). The following theorem describes
the geometric properties of V(l1 ) and V(l2 ).
RHIT Undergrad. Math. J., Vol. 15, No. 1
Page 13
Theorem 3.2. Let D ∈ C[x, y] define an irreducible cubic curve through the point p and let
Q define an irreducible quadratic such that Ip (D, Q) ≥ 5. In the form
D = T 2 l1 + Ql2
afforded by the previous theorem, V(l1 ) contains both p and the sixth point of intersection of
V(Q) ∩ V(D). Furthermore, V(l2 ) is a tangent line to V(D) at the third point of intersection
of V(T ) ∩ V(D).
2
l2
T
1
Q
l1
s
0
p
q2
-1
q1
C
-2
-2
-1
0
1
2
Proof. Since Ip (D, Q) ≥ 5, the notion of a sixth point of intersection of V(Q) ∩ V(D) is well
defined and may or may not be the point p. Let q1 denote this point. Recall that l1 (p) = 0,
so if it is the case that q1 = p, the result follows trivially. Suppose then, that q1 is distinct
from p and consider the following evaluation
D(q1 ) = T (q1 )2 l1 (q1 ) + Q(q1 )l2 (q1 )
0 = T (q1 )2 l1 (q1 ).
We know that Ip (T, Q) = 2 since T defines the tangent line of Q at p and by Bézout, these
curves can only intersect twice including multiplicity. It follows that since q1 is distinct from
p, that T (q1 ) 6= 0 so l1 must vanish at q1 .
Similarly, since Ip (D, T ) ≥ 2, it makes sense to talk about the “third” point of intersection
of V(T ) ∩ V(D). We will call this point q2 . An evaluation at q2 gives us
D(q2 ) = T (q2 )2 l1 (q2 ) + Q(q2 )l2 (q2 )
0 = Q(q2 )l2 (q2 ).
Page 14
RHIT Undergrad. Math. J., Vol. 15, No. 1
However, if Q(q2 ) = 0 then the curves V(T ) and V(Q) would intersect at the point q2 in
addition to intersecting at p with a multiplicity of 2, totalling 3 intersection points. This
would force V(Q) to contain the line V(T ) by Bézout, a contradiction to the irreducibility
of Q. Therefore Q(q2 ) 6= 0 which means that l2 (q2 ) = 0 and further implies that q2 6= p since
Q(p) = 0. It then follows that
Iq2 (D, l2 ) = Iq2 (T 2 l1 + Ql2 , l2 ) = Iq2 (T 2 l1 , l2 )
= 2Iq2 (T, l2 ) + Iq2 (l1 , l2 )
= 2 + Iq2 (l1 , l2 )
From this, it follows that Iq2 (D, l2 ) ≥ 2. This implies that V(l2 ) is tangent to V(D) at q2
so long as q2 is a smooth point of V(D). If q2 is not smooth on V(D), then Iq2 (T, D) ≥ 2.
However, it is already the case that Ip (T, D) ≥ 2 and since q1 6= p, this would lead to a
contradiction since a line can only intersect a cubic at three points including multiplicity.
Hence, q1 must be a smooth point and thus V(l2 ) is tangent to V(D) at q2 .
3.1
Case 1: Smooth, Non-Flex Points On Cubics
Throughout this section we will let C ∈ C[x, y] be an irreducible polynomial that defines a
cubic curve that is smooth at the point p = (0, 0) where p is not a flex point of V(C). As
such, by Lemma 2.24, we know that there exists an osculating conic of V(C) at p. We will let
Q denote a defining polynomial of this osculating conic and T denote a defining polynomial
of the tangent line of V(C) at p. By Lemma 2.26, we know that both V(T ) and V(Q) are
unique. Of course there are infinitely many equations of these curves, all of which differ by
multiplication of a constant. In each instance we simply fix one such equation.
Furthermore, by Theorem 3.1, we know we can write C = T 2 l1 + Ql2 . From this point on
we will use the notation l1 and l2 to refer precisely to these linear terms. We will also refer
to q1 and q2 as defined in Theorem 3.2 as the third point of intersection of V(T, C) and the
sixth point of intersection of V(Q, C) respectively.
Our goal now is to investigate whether we can find another cubic curve that intersects
V(C) to degree 9 at p.
Lemma 3.3. Let D ∈ C[x, y] define an irreducible curve. Then Ip (C, D) = 9 if and only
if D can be written as D = T 2 l3 + Ql4 where l3 , l4 ∈ C[x, y] are linear polynomials and
V(Q) = V(l1 l4 − l2 l3 ). Furthermore, if Ip (C, D) = 9, then p is not a sextactic point of V(C).
Proof. We know that Ip (C, Q) ≥ 5, so if Ip (C, D) = 9, then transitively by Lemma 2.3,
Ip (D, Q) ≥ 5 which fulfills the necessary conditions in Theorem 3.1 for D to be written as
D = T 2 l3 + Ql4 .
Suppose that V(C) and V(D) intersect only at a point p with multiplicity 9. Then
RHIT Undergrad. Math. J., Vol. 15, No. 1
Page 15
equivalently
9 = Ip (C, D) = Ip (T 2 l1 + Ql2 , T 2 l3 + Ql4 )
= Ip (T 2 l1 + Ql2 , T 2 l3 l2 + Ql4 l2 )
= Ip (T 2 l1 + Ql2 , T 2 l3 l2 − T 2 l4 l1 )
= Ip (T 2 l1 + Ql2 , −T 2 (l1 l4 − l2 l3 ))
= Ip (C, −T 2 ) + Ip (C, (l1 l4 − l2 l3 ))
= 4 + Ip (C, (l1 l4 − l2 l3 )).
The above equality holds if and only if Ip (C, l1 l4 − l3 l2 ) ≥ 5. But V(l1 l4 − l3 l2 ) is a
quadratic, so in order for it to intersect V(C) with degree 5, it must be the osculating
quadratic of V(C), namely V(Q).
Furthermore, if p was a sextactic point of V(C), we would have that Ip (C, Q) = Ip (C, l1 l4 −
l2 l3 ) = 6 which would imply that Ip (C, D) = 10. This is a contradiction by Bézout’s Theorem
since C and D are polynomials of degree 3 so therefore, p cannot be a sextactic point of
V(C).
Lemma 3.4. If there is one cubic curve defined by D ∈ C[x, y] such that Ip (C, D) = 9, then
there are infinitely many. Furthermore, every such cubic other than D itself is of the form
C + αD for some α ∈ C∗ .
3
2
1
0
-1
-2
-3
-3
-2
-1
0
1
2
3
Proof. It follows directly from the properties of intersection multiplicity that Ip (C, D) = 9
implies Ip (C, C + αD) = 9 for α 6= 0.
RHIT Undergrad. Math. J., Vol. 15, No. 1
Page 16
We next show that every point in the plane is contained by one of the curves in this
family. Let s be an arbitrary point in the plane. If s ∈ V(D) or s ∈ V(C) we are done.
Suppose not, then letting α = −C(s)/D(s) yields a curve passing through s with α 6= 0.
Let E be the defining polynomial of any curve intersecting C to degree 9 at p. Pick
any point s ∈ V(E), s 6= p. By our previous observation there must exist some curve F in
our infinite family passing through s. However, E must intersect F to degree 9 at p and to
degree at least 1 at s which is a contradiction of Bézout’s Theorem unless they describe the
same curve. Therefore this family contains all such curves.
Theorem 3.5. There exists a cubic curve which intersects V(C) at p with degree 9 if and
only if V(l1 ) is tangent to V(C) at q1 .
3
2
T
1
Q
l1
0
p
q1
-1
C
q2
-2
-3
l2
-3
-2
-1
0
1
2
3
Proof. Recall that p, q1 ∈ V(l1 ) and V(l2 ) is a tangent line of V(C) at q2 .
Suppose that V(l1 ) is tangent to V(C) at q1 . Then it must be the case that V(l1 , C) =
{p, q1 } since V(l1 ) and V(C) intersect at 3 points including multiplicity, two of which are q1
and the other p.
Let s denote the point of intersection of V(l1 ) and V(l2 ). An evaluation of the equation
C = T 2 l1 + Ql2 at s shows that C(s) = 0 so s ∈ V(C).
Since s ∈ V(l1 , C) we must have that s is either p or q1 . Either way, s ∈ V(l1 , Q) so
s = V(l1 , l2 ) ⊂ V(Q). According to the ideal-variety correspondence, it follows that
Q ∈ hl1 , l2 i,
RHIT Undergrad. Math. J., Vol. 15, No. 1
Page 17
so we can write
Q = l1 A − l2 B
for some linear polynomials A and B. Then it follows directly from Lemma 3.3 that the
curve
D = T 2 B + QA,
intersects V(C) at p with degree 9.
Conversely, suppose that V(l1 ) is not tangent to V(C) at q1 . Recall that if Ip (C, D) = 9,
then p cannot be a sextactic point of C. This is precisely the same thing as saying that
p 6= q1 . Since p, q1 ∈ V(l1 , C), it makes sense to discuss the third point of V(l1 , C) which we
will call r.
We know r 6= q1 since otherwise, this would imply that V(l1 ) is tangent to V(C) at q1 .
We know r 6= p since otherwise V(l1 ) = V(T ) and it would follow that
Ip (C, Q) = Ip (T 3 + Ql2 , Q) = 3Ip (T, Q) = 6.
Which is equivalent to saying p is sextactic on V(C), a contradiction. Finally, we conclude
that since r ∈ V(C) but r 6∈ {p, q1 } = V(C, Q) we must have that r 6∈ V(Q). Evaluating the
usual form for C at r yields
C(r) = T 2 (r)l1 (r) + Q(r)l2 (r)
0 = Q(r)l2 (r)
Since r 6∈ V(Q), it must be the case that l2 (r) = 0.
Suppose now toward a contradiction that D = T 2 l3 + Ql4 defines a cubic curve such that
Ip (C, D) = 9. Then by Lemma 3.3 it follows that
Q = l1 l4 − l2 l3 ,
but since l1 (r) = 0 and l2 (r) = 0, this would imply that
Q(r) = 0,
which is contradiction since r 6∈ V(Q).
Therefore, if V(l1 ) is not tangent to V(C) at p, we may conclude that there does not
exist any cubic which intersects V(C) at p with degree 9.
Remark 3.6. Note that if D ∈ C[x, y] defines a cubic curve that intersects V(C) at p with
multiplicity 9, it must be irreducible. Otherwise, it would decompose into three lines, or a
line and a conic. Either way, by properties of intersection multiplicity, the line would have
to intersect V(C) with multiplicity 3 at p, contradicting the fact that p is not a flex point.
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RHIT Undergrad. Math. J., Vol. 15, No. 1
Lemma 3.7. If there is an infinite family of cubics that intersect V(C) at p to degree 9,
precisely one of those cubics has a singularity at p. In particular, the singularity is a node
and the singular curve may be written as
N = T 3 + QT 0 ,
where T 0 defines the second tangent of V(N ) at p and Ip (N, Q) = 6.
Proof. Let D ∈ C[x, y] define an irreducible cubic curve such that Ip (C, D) = 9. It follows
that Ip (T, D) ≥ 2 and thus V(T ) is the tangent line of V(D) at p. Thus, the linear terms
of C and D must differ merely by a constant. Since multiplying a polynomial by a constant
does not change its solutions, we may assume that C and D have the same linear term.
Recall that if Ip (C, D) = 9, then Ip (C, C + αD) = 9 for any α ∈ C∗ . Let N = C + (−1)D.
Since C and D have the same linear term, it follows that N has no linear term, and thus is
singular at the point p. Since Ip (C, Q) = 5 and Ip (C, N ) = 9 it follows that Ip (N, Q) ≥ 5 so
we may write N = T 2 l3 + Ql4 where l4 vanishes at p since N is singular.
Next, we will show that l3 and l4 both define tangent lines to V(N ) at p that are distinct.
Consider the following
Ip (l4 , N ) = Ip (l4 , T 2 l3 + Ql4 )
= Ip (l4 , T 2 l3 )
= 2Ip (l4 , T ) + Ip (l4 , l3 )
=3
Ip (l3 , N ) = Ip (l3 , T 2 l3 + Ql4 )
= Ip (l3 , Ql4 )
= Ip (l3 , (l1 l4 − l2 l3 )l4 )
= Ip (l3 , l1 l42 − l2 l3 l4 )
= Ip (l3 , l1 l42 )
= Ip (l3 , l1 ) + 2Ip (l3 , l4 )
=3
Note that l3 6= l4 since otherwise Q = l1 l4 − l2 l3 would be reducible. Hence, l3 and l4 define
distinct tangent lines to N at p which means N has a node at p. By the remark preceeding
this lemma, N = T 2 l3 + Ql4 is irreducible so V(l4 ) 6= V(T ). Therefore, since nodes only have
two tangents, we conclude that V(l3 ) = V(T ) and that l4 must define the second tangent
of V(N ) at p. Thus, after multiplication by a constant, we may write N = T 3 + QT 0 and
consider Ip (N, Q).
Ip (N, Q) = Ip (T 3 + QT 0 , Q)
= Ip (T 3 , Q)
= 3Ip (T, Q)
= 6.
RHIT Undergrad. Math. J., Vol. 15, No. 1
Page 19
Finally, to see that this node is unique, notice that every cubic which intersects V(C)
with multiplicity 9 at p is of the form C + αD and that the only choice of α that makes the
linear term of C + αD vanish is α = −1.
Theorem 3.8. If there is an infinite family of cubics that intersect V(C) to ninth order
at p, then the members of this family can be described by a one to one correspondence with
points on V(Q).
3
2
Dt
T
l4
1
Q
l1
0
qt
p
q1
l3
-1
q3
C
q2
-2
-3
l2
-3
-2
-1
0
1
2
3
Proof. Define the point q3 to be the point at which V(Q) and V(l2 ) intersect other than q1
and let qt be any point on V(Q). Define At to be an equation of the line pqt and Bt to be
an equation of the line q3 qt . Let us examine the set V(At l2 , Bt l1 ). We have by definition
V(At , Bt ) = {qt }
V(At , l1 ) = {p}
V(l2 , Bt ) = {q3 }
V(l2 , l1 ) = {q1 }.
So equivalently, we have that V(at l2 , Bt l1 ) = {p, q1 , q3 , qt } ⊂ V(Q). By Max Noether’s
Fundamental Theorem, this implies that we can write
Q = αl1 Bt − βl2 At ,
where α, β ∈ C. But since multiplying by a scalar does not change the vanishing of a
polynomial, we can define A0t = βAt and Bt0 = αBt to be new equations for the lines defined
by At and Bt .
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RHIT Undergrad. Math. J., Vol. 15, No. 1
Now since we have Q = l1 Bt0 − l2 A0t , by a previous theorem we have that the curve defined
by
Dt = T 2 A0t + QBt0 ,
intersects the curve V(C) to degree nine at p.
Conversely, suppose that we have some Dt such that V(Dt ) intersects V(C) to degree
nine at p. We can write D = T 2 A + QB where A and B are the analogous constructions
of l1 and l2 in C = T 2 l1 + Ql2 . We can then consider the point of intersection of V(A)
and V(B) and name it s. Setting qt in the above construction to be equal to s yields that
V(D) = V(Dt ). So the points on V(Q) are in a one to one correspondence with the elements
of the family of cubics that intersect C to degree 9 at p.
3.2
Constructing Families from an Irreducible Nodal Cubic
We know that in the family of cubic curves constructed in the previous section, there is one
particular cubic curve that has a node at p. It is worthwhile to ask the inverse question:
Does every nodal cubic give rise to such a family of smooth cubics? Are these families in a
one to one correspondence with nodal cubics? Or can one nodal cubic be the unique singular
curve of more than one family? We will explore questions such as this now.
Throughout this section, we will suppose that N ∈ C[x, y] defines an irreducible cubic
curve with a node at p = [0 : 0 : 1].
Remark 3.9. Recall that by Lemma 2.25, there exists two osculating conics to V(N ) at p.
Each conic corresponds to a different tangent line to V(N ) at p so by Theorem 3.1 we may
write N in either of the following forms.
N = T13 + Q1 T2
N = T23 + Q2 T1
Lemma 3.10. Let N = T13 + Q1 T2 = T23 + Q2 T1 . There exists two smooth cubic curves
defined by C1 , C2 ∈ C[x, y] such that Ip (N, C1 ) = 9 and Ip (N, C2 ) = 9 but Ip (C1 , C2 ) ≤ 9.
Furthermore, these cubic curves may be written as
C1 = T12 l1 + Q1 l2 ,
and
C2 = T22 l3 + Q1 l4 .
Proof. By Theorem 3.3, we only need to show that there exists l1 and l2 such that V(Q1 ) =
V(T1 l2 − T2 l1 ) in order to prove that
C1 = T12 l1 + Q1 l2
defines a smooth cubic curve with Ip (N, C1 ) = 9.
RHIT Undergrad. Math. J., Vol. 15, No. 1
Page 21
We will show that such l1 and l2 exists by construction.
Let q3 be the second point of intersection of V(Q1 ) and V(T2 ) and fix some q4 ∈ V(Q1 ).
Let l10 be an equation of the line connecting p and q4 and let l20 be an equation of the line
connecting q3 and q4 .
Considering the variety V(T1 l2 , T2 l1 ) we see that
V(T1 , T2 ) = {p}
V(T1 , l10 ) = {p}
V(l20 , T2 ) = {q3 }
V(l20 , l10 ) = {q4 },
so V(T1 l20 , T2 l10 ) ⊆ V(Q1 ) so by the ideal variety correspondence we have that
Q1 ∈ hT1 l20 , T2 l10 i.
It then follows that Q = α(T1 l20 − T2 l10 ) = T1 (αl20 ) − T2 (αl10 ). Setting l1 = αl10 and l2 = αl20
we see that l1 and l2 are defining equations of the same lines defined by l1 and l2 . Thus, the
linear polynomials l1 and l2 satisfy the conditions of 3.3 so setting C1 = T12 l1 + Q1 l2 yields
the result that Ip (C1 , N ) = 9.
We can repeat the same construction after fixing Q2 instead of Q1 to see that C2 =
T12 l3 + Q1 l4 yields the result of Ip (C2 , N ) = 9.
However, observe that if Ip (C1 , C2 ) = 9 then Ip (C1 , Q2 ) = 5 which would imply that Q1
and Q2 defined the same conic. This is a contradiction since we chose Q1 and Q2 to define
the two unique conics that intersect V(N ) to degree 6 at p.
Lemma 3.11. If there exists an irreducible cubic curve defined by C ∈ C[x, y] such that
Ip (C, N ) = 9, then there is an infinite family of such cubic curves which all intersect each
other at p with degree 9.
Proof. Let C ∈ C[x, y] define an irreducible cubic curve such that Ip (C, N ) = 9. Consider
the curve defined by Ct = C + tN . It directly follows that Ip (Ct , N ) = 9 and Ip (Ct , C) = 9
for all t ∈ C.
Theorem 3.12. For every nodal cubic N there are two families of cubic curves EQ1 and EQ2
such that Ip (A, B) = 9 for every A, B ∈ EQi and if A ∈ EQ1 and B ∈ EQ2 then Ip (A, B) = 9
if and only if A = B = N .
Proof. By Lemma 3.10 there are at least two cubic curves defined by C and D such that
Ip (C, N ) = 9, Ip (D, N ) = 9 and Ip (C, D) < 9. However, by Lemma 3.11 we know that there
are infinitely many Ct and Dt such that Ip (C, Ct ) = 9, and Ip (D, Dt ) = 9. However, since
Ip (C, D) < 9, we may conclude from Lemma 2.3 that Ip (Ci , Dj ) < 9 for all smooth Ci and
Dj . Note that all Ci are smooth except N and all Dj are smooth except N so the only
exception to Ip (Ci , Dj ) < 9 is when Ci = Cj = N .
Page 22
RHIT Undergrad. Math. J., Vol. 15, No. 1
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Introduction to Computational Algebraic Geometry and Commutative Algebra, 3/e (Undergraduate Texts in Mathematics). Springer-Verlag New York, Inc., New York, NY,
USA, 2007.
[3] William Fulton. Algebraic Curves: An Introduction to Algebraic Geometry. AddisonWesley, 1989.
[4] Alwaleed Kamel and M. Farahat. On the moduli space of smooth plane quartic curves
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